Leaving Groups in Substitution and Elimination Reactions (vid 1 of 2) by Leah4sci

Leah here from Leah4Sci.com and in this video,
I will show you how to analyze the effects of leaving groups in SN1, SN2, E1, and E2
reactions. But first, what is a leaving group? A leaving group as the name implies is the
atom or molecule that leaves either by getting eliminated or kicked out in a substitution
and elimination reaction. In a substitution reaction, when you react
the carbon chain containing the leaving group with a nucleophile, your product is going
to have the carbon chain with a nucleophile attached where the leaving group used to be
and the leaving group floating around the solution. In an elimination reaction, your carbon containing
the leaving group is going to react with a base but instead of the base replacing the
leaving group, your leaving group as well as your beta hydrogen will be gone and there
will be a pi bond where those two atoms used to be. As you're analyzing the effect of leaving
group on a reaction, don't forget that you also have to look at the following: the alkyl
chain, the attacking molecule, and the solvent in determining between substitution and elimination
and in determining between the one type, meaning, SN1, E1 or the two-type SN2 or E2 reactions
because it is these factors that will usually determine between substitution and elimination
but the leaving group will help us determine between the one-type and the two-type and
also, the speed of the reaction. Let's analyze the one-type reaction given
that SN1 and E1 start out the same way where the leaving group grabs a lone pair of electrons
that connected to the carbon chain and disappears off of the molecule. This leaves you with a carbon chain that has
a positive charge and a leaving group that has the extra lone pair of electrons. Assuming all factors allow for the carbocation
to form, the speed of this reaction will be determined by the leaving group's willingness
to leave. In the two-type reaction, the leaving group
is kicked out so we don't analyze it as closely because it's typically not given a choice. Picture this scenario. You're watching TV with your younger sibling
and you're sitting in your favorite chair. You get up to grab a cup of water and your
sibling steals your chair. Now, I don't know about you but if I see my
younger sister sitting in my chair, I kick her out. I'm bigger, I'm stronger, and that's my chair. In other words, when I left, I was not happy
to be gone and because I wanted to come back, that would make me a bad leaving group. My younger sister on the other hand, didn't
have a choice because I kicked her out it didn't matter if she was a good leaving group
or not because I was stronger and took that choice away. In other words, when you analyze the ability
of a leaving group, don't look so much at the leaving group's willingness to leave as
much as the ability of the leaving group to stay gone. As a rule of thumb for leaving groups, a good
leaving group is a weak base. In other words, it's the conjugate base of
a strong asset. But often, you're not given a pKa Table to
determine the asset's strength and so you have to understand how to actually analyze
the atoms or group of atoms to determine if your reaction will proceed quickly, slowly,
or not at all. So how do you know if a leaving group is happy
to stay gone? Let's use a real example and analyze what
happens to the charge of that leaving group. We'll start with a molecule 2-bromobutane
where in this case, bromine is the leaving group. If we do a quick formal charge on bromine,
it should have 7 valence electrons and it does have 7 electrons directly attached giving
it a formal charge of zero. As a leaving group, bromine grabs the 2 red
electrons breaking the bond between bromine and carbon and takes it as it leaves the carbon
chain. In addition to the carbocation that is formed,
we now also have a bromine with a full octet with its three initial lone pairs of electrons
plus those two red electrons that it took from the bond to carbon. A quick formal charge calculation shows me
that bromine should have had 7 valence electrons, 7 minus 8 gives me -1 and so, I have a negative
charge on bromine. Now the question you have to ask yourself,
is bromine capable of holding that negative charge? When it comes to the ability of an atom to
hold the charge, size matters. Look at the trend of the halogens on the Periodic
Table. You'll see that we have fluorine then chlorine
then bromine then iodine telling me that fluorine is the smallest and iodine is the largest. If I place a negative charge on iodine or
iodide, the negative charge will be well-distributed over the large atom making it more capable
of holding that charge and therefore more stable. Compare this to a negative chlorine or chloride. The atom is smaller, the charge is less distributed
or more concentrated making this a little bit more reactive and therefore a little bit
less stable. Well, chlorine is considered an OK leaving
group. Iodine is considered a very good leaving group
because it can hold that negative charge better. Now, let's take a look at oxygen as a leaving
group. Oxygen which shows up on the second period
is a rather small and highly electronegative atom. When oxygen grabs the two bonding electrons
to break away from the carbon, once again we wind up with a carbocation and a negative
oxygen in solution. Unlike the halogens, oxygen which is so small,
highly electronegative with concentrated charge is considered to be a very strong base or
strong nucleophile and therefore, if you manage to kick it out, it won't stay gone but instead
will use its negative electrons to attach right back and reverse the reaction. But oxygen isn't always a terrible leaving
group. The reason OH- was not happy leaving is because
the oxygen was left carrying that entire burden of charge. However, if you can have an oxygen as a leaving
group that shares the charge with another atom, it will be a lot happier, a lot more
stable and therefore less likely to come back and attack. Say I have another molecule that has oxygen
as a leaving group. But this oxygen is attached to a carbonyl
and then a CH3. You'll recognize that this is an acetate portion. Meaning, the conjugate base of acidic acid
which is a decent acid and when the bond between carbon and oxygen breaks, it's not only oxygen
that is left carrying the charge. Let's analyze just the leaving group. The negative charge sits on the oxygen atom
but that oxygen is bound to a partially positive carbonyl carbon bound to a partially negative
carbonyl oxygen. And so, this negative oxygen can use its electrons
to attack towards the carbonyl position breaking the pi bond between carbon and the second
oxygen. To show the alternate resonance form, we start
with a skeleton and now let's see what happened with the electrons. We have the black pair of the electrons that
formed a carbonyl between that left oxygen and carbon. And we'll show these pi electrons as being
purple breaking on to the oxygen. As you can see, the negative charged moved
on to the other oxygen but in reality, it's constantly going back and forth. Therefore, we don't have one oxygen holding
a negative charge but instead we have two oxygens sharing that negative charge since
half of a negative charge is less of a burden on a full charge. This is stable molecule and therefore a good
leaving group. But what do you do when you really have to
kick out your OH? Well, you bribe it of course. Be sure to join me in the next video 2 of
2 of the leaving group series where I show you how to bribe this OH group using either
an acid catalyst or a tosyl group. Are you struggling with organic chemistry? Are you looking for information to guide you
through the course and help you succeed? If so, download my ebook, 10 Secrets to Acing
Organic Chemistry using the link below or visit Leah4Sci.com/OrgoSecrets. For information regarding online tutoring,
visit Leah4Sci.com/OrgoTutor. If you enjoyed this video, please give it
a thumbs up and even share it with a friend or two. If you have any questions regarding this video,
leave a comment below or contact me through my Facebook page at Facebook.com/Leah4Sci. There will be many related videos posted over
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14 thoughts on “Leaving Groups in Substitution and Elimination Reactions (vid 1 of 2) by Leah4sci

  1. Obviously you are bigger and stronger. But don't kick out your sweet cute sister😁😆😅😂love her. I also love your vedios

  2. Thank you so much, ma'am Leah! I have an org chem exam coming up, and your videos are a big help!

  3. Nice video. I have a question that's been going around my mind that if F is more electronegative than I,then F should be happy to hold the negative charge as compared to I.Your explanation is good regarding the volume occupied but I'd like to know why electronegativity doesn't have a good part.Thank you in advance.

  4. I am very much dependent on your videos to get " A plus" in my two classes. I am hopeful that I will end up not only good marks but also good understanding of the concepts.

  5. a small question ,why good leaving groups are weak bases . I have gone through your videos they were amazing thanku

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